Exercise : Let G finite group, $\alpha$ automorphism of G

Question

This morning during an exam, it was given to me the following exercise that i was unable to solve, and i would like to know at least how should be solved.

Maybe it was my bad, but it was the first time that i saw an Automorphism and i didn't really much more that if an isomorphism $f : G \longrightarrow G$

Let $G$ be a finite group and $\alpha$ an Automorphism of $G$ such that $\alpha(x)=x$ if and only if $x =e_G$

$(a)$ Prove that for every $g \in G$ exists $x \in G$ such that $g=x^{-1}\alpha(x)$.

$(b)$ Prove that if it's true that $\alpha(\alpha(x))=x$ for every $x \in G$, then $\alpha(g)= g^{-1}$ for every $g \in G$.

Your help would be amazing,

Thank you anyway.

Answer 1

Consider $f: G \to G$ defined by $f(x) = x^{-1}\alpha(x)$. Part a) asks to show that $f$ is surjective. Since $|G|$ is finite, it suffices to show that $f$ is injective. So suppose that $f(x) = f(y)$. Then $$x^{-1}\alpha(x) = y^{-1}\alpha(y),$$ so multiplying on the left by $y$ and on the right by $\alpha(x)^{-1} = \alpha(x^{-1})$ shows $$yx^{-1} = \alpha(y)\alpha(x)^{-1} = \alpha(yx^{-1}).$$ But since $\alpha(z) = z$ if and only if $z$ is equal to the identity, this implies that $yx^{-1}$ is the identity, i.e. $x=y$.

To solve part b), observe that if $g \in G$, then $g = x^{-1}\alpha(x)$ for some $x \in G$ by part a). Then $$\alpha(g) = \alpha(x)^{-1} \alpha(\alpha(x)) = \alpha(x)^{-1}x = (x^{-1}\alpha(x))^{-1} = g^{-1},$$ as desired.

Answer 2

For part $a$ suppose $x^{-1}\alpha (x)=y^ {-1}\alpha(y)$ and conclude $xy^{-1}=\alpha(xy^{-1})$. Hence $x=y$, since the group is finite this means $x^{-1}\alpha(x)$ covers the group.

For part $b$ notice $\alpha(x^{-1}\alpha(x))=\alpha(x^{-1})x=(x^ {-1}\alpha(x))^{-1}$.

And since $x^{-1}\alpha(x)$ covers the whole group we are done.

Answer 3

Let $a=\alpha$ be the given automorphism.

(a) Let us consider the appropiate map (of sets), $G\to G$, $x\to f(x):= x^{-1}a(x)$. Let us show the injectivity of this map. Start with $x,y\in G$, and assume $f(x)=f(y)$. Then we can successively rewrite equivalently: $$ \begin{aligned} f(x)&=f(y)\ ,\\ x^{-1}a(x) &= y^{-1}a(y)\ ,\\ yx^{-1}a(x) &= a(y)\ ,\\ yx^{-1} &= a(y)\, a(x)^{-1}\ ,\\ yx^{-1} &= a(y\, x^{-1})\ , \end{aligned} $$ so the element $yx^{-1}$, invariated by $a$ is the neutral element. This implies $x=y$.

We have thus shown the injectivity.

Since $G$ is finite, the injectivity of the self-map $f$ implies its surjectivity. It is exactly what we have to show in (a). Moreover the $x$ is unique.

(b) Fix some $g$ in $G$. We find an $x$ as in (a). Then: $$ \begin{aligned} a(g) &= a(x^{-1}a(x))\\ &= a(x^{-1})\, a(a(x))\\ &= a(x^{-1})\, x\\ &= a(x)^{-1}\, x\\ &= (x^{-1}a(x))^{-1}\\ &=g^{-1}\ . \end{aligned} $$ Note: In this case, this also implies that the group $G$ is commutative.